In a certain A.P. the 24th term is twice the 10th term.

Solution:

Given:

$a_{24}=2 a_{10}$

$\Rightarrow a+(24-1) d=2[a+(10-1) d]$

$\Rightarrow a+23 d=2(a+9 d)$

$\Rightarrow a+23 d=2 a+18 d$

$\Rightarrow 5 d=a \quad \ldots$ (i)

To prove:

$a_{72}=2 a_{34}$

LHS : $a_{72}=a+(72-1) d$

$\Rightarrow a_{72}=a+71 d$

$\Rightarrow a_{72}=5 d+71 d \quad(\operatorname{From}(\mathrm{i}))$

$\Rightarrow a_{72}=76 d$

RHS : $2 a_{34}=2[a+(34-1) d]$

$\Rightarrow 2 a_{34}=2(a+33 d)$   (From (i))

$\Rightarrow 2 a_{34}=2(38 d)$

$\Rightarrow 2 a_{34}=76 d$

$\therefore \mathrm{RHS}=\mathrm{LHS}$

Hence, proved.